if f(x)=2x-3 and g(x)=2x^2+1 find A.f(g(2)) B. g(f(x)) C. does f(x) have and inverse if so find the inverse of f(x)

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if f(x)=2x-3 and g(x)=2x^2+1 find A.f(g(2)) B. g(f(x)) C. does f(x) have and inverse if so find the inverse of f(x)

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f surely has an inverse because it is a line and therefore one to one.
f says 'multiply by 2 then subtract 3' so f inverse would say to do the opposite things in the opposite order: add 3 and divide by 2 so \[f^{-1}(x)=\frac{x+3}{2}\]
if this is confusing, rewrite \[f(x)=2x-3\] as \[y = 2x-3\] then switch x and y (because that is what the inverse does) to get \[x=2y-3\] and solve this for x: \[x=2y-3\] \[x+3=2y\] \[\frac{x+3}{2}=y\] and y is your inverse.

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Other answers:

is that the answer for C
yes. was it clear?
ok so whats the answers for a and b
\[f(g(x))=f(2x^2+1)=2(2x^2+1)-3\]
i guess there is a little algebra to do now: \[2(2x^2+1)-3=4x^2+2-3=4x^2-1\]
but it was f(g(2))
is it clear what i did? first write \[f(g(x))\] then replace \[g(x)\] by \[2x^2+1\] and then rewrite f replacing \[x\] by \[2x^2+1\]
oh well if it is \[f(g(2))\] then since \[f(g(x))=4x^2-1\] then \[f(g(2))=4(2^2)+1=17\]
typo sorry. \[4(2^2)-1=15\]
or you could say \[g(2)=2(2^2)+1=9\] and \[f(9)=2\times 9 - 1=15\]
ok
\[g(f(x))=g(2x-3)=2(2x-3)^2+1\]
this requires more algebra: \[2(2x-3)^2+1=2(2x-3)(2x-3)+1=2(4x^2-6x+9)+1\] \[=8x^2-12x+18+1=8x^2-12x+19\] if my algebra is correct.
ok for a how come its 4(2^2)-1
and not 2(2^2) +1

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